There is a proof online which uses the following:
$$1 + n + {n\choose 2} + \ldots + {n \choose m} \leq n^m \quad \text{for } n,m\geq 2$$
Is there a name of this bound? Or a short proof anywhere? The ones given here are more 'advanced' since I believe this one is really loose.
The step is used at section 2.3 of this file.
Hint: Following from your comment, $${n \choose m} =\frac{n!}{m!(n-m)!}=\frac{n(n-1)\cdots(n-m+1)}{m(m-1)\cdots1}.$$ We can assume that $m \ge n/2$ (just for the next step!) (otherwise consider ${n \choose n-m}$) and when this is the case $${n \choose m} = \frac{n(n-1)\cdots(n-m+1)}{m(m-1)\cdots1} = \frac{n}{m} \cdot \frac{(n-1)\cdots(n-m+1)}{(m-1)\cdots1} \le 2 n^{m-1}.$$ This gives $$n^{m-1} + {n\choose m} \le 3n^{m-1}$$ and so the conclusion holds for $n \ge 3$. It's easy to check the $n=2$ case.
Hope this helps!